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This is true for finite-dimensional spaces: the diagonal operators on a finite dimensional complex vector space form contain a dense open set and the nondiagonalizable operators have measure 0.

To be precise, let $T$ be an operator on a complex Banach space $X$ which is not finite-dimensional. For each $\lambda \in \mathbb{C}$, let $V_\lambda \subseteq X$ be the subspace $\mathrm{ker} (\lambda I - T)$ on which $T$ acts by the scalar $\lambda$. Say that $T$ is diagonalizable if $\sum_\lambda V_\lambda$ is dense in $X$. Or provide a better definition if this one is deficient!

I want to know to what extent the "typicality" of diagonalizable operators carries over to infinite dimensions. Are the diagonal operators dense? Are they open (or do they contain an open set)? Are they comeagre? Of course, this will probably depend on the Banach space and the operator topology. I suppose it's natural to consider just bounded operators, although I'd be interested in results about unbounded operators, too. Also, the question makes perfect sense for any topological vector space; I'm interested in non-Banach spaces, too.

I asked this question on math stackexchange, and Mateusz Wasilewski pointed out that the Weyl - von Neumann - Berg theorem shows that on separable Hilbert space, the "orthogonally diagonalizable" operators (where the $V_\lambda$ are required to be orthogonal) are dense among normal operators (in the norm topology).

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    $\begingroup$ Diagonalizable operators are not even open in finite-dimensional vector spaces, so that one at least is unlikely. $\endgroup$
    – Will Sawin
    Commented Apr 8, 2015 at 17:15
  • $\begingroup$ Ah... I see the argument I had in mind only shows that diagonal operators with distinct eigenvalues form an open dense set -- Thanks. $\endgroup$
    – Tim Campion
    Commented Apr 8, 2015 at 17:19
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    $\begingroup$ Just a comment, given the answer below and the ensuing comments: you will always have problems as soon as there exists something like a one-sided shift, i.e. a map which is injective with closed range but not surjective -- for such operators, there are points in the spectrum that are not approximate eigenvalues, and these are an obstruction to approximation by operators where every point is an approximate eigenvalue (this class including diagonalizable ones, I think) $\endgroup$
    – Yemon Choi
    Commented Apr 19, 2015 at 18:21

1 Answer 1

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Consider the right shift $R([x_1, x_2, \ldots]) = [0, x_1, x_2, \ldots]$ on $\ell^2$. I claim the open ball of radius $1/2$ about $R$ contains no diagonalizable operators.

Let $e_1 = [1,0,\ldots]$. Suppose $B$ is an operator with $\|B\| < \epsilon < 1/2$. For any $x$ we have $\|(R + B)x \| \ge (1-\epsilon) \|x\|$, and so $$ \langle e_1, (R+B) x\rangle = \langle e_1, B x \rangle \le \|B x\| \le \epsilon \|x\| \le \dfrac{\epsilon}{1-\epsilon} \|(R+B)x\| $$ Since $\epsilon/(1-\epsilon) < 1$, this implies that $e_1$ is not in the closure of $\text{Ran}(R+B)$. Now any eigenvector of $R+B$ is in $\text{Ran}(R+B)$ (note that $0$ is not an eigenvalue because $\|(R+B)x\| \ge (1-\epsilon)\|x\|$), so the span of the $V_\lambda$ for $R+B$ is not dense.

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  • $\begingroup$ Very nice. I'll refrain from accepting this answer for a little while only because it's not immediately obvious how to extend the argument to more general Banach spaces. $\endgroup$
    – Tim Campion
    Commented Apr 8, 2015 at 20:53
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    $\begingroup$ There are infinite dimensional Banach spaces on which every bounded linear operator is of the form $\lambda I + K$ with $K$ a compact operator. On such a space the diagonal operators are dense. $\endgroup$
    – Bill Johnson
    Commented Apr 8, 2015 at 21:31
  • $\begingroup$ @Bill As might be clear from the question, I know very little functional analysis. I believe that the spectral theory of compact operators shows that every compact operator on a Hilbert space is diagonalizable: is this true for all Banach spaces? If so, then in the spaces you're talking about, every operator is digaonalizable, right? $\endgroup$
    – Tim Campion
    Commented Apr 8, 2015 at 22:00
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    $\begingroup$ No, that's not even true in finite dimensions (where every operator is compact). $\endgroup$
    – Robert Israel
    Commented Apr 8, 2015 at 23:11
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    $\begingroup$ The Argyros-Haydon spaces I mentioned have Schauder bases, so the finite rank operators are dense in the compact operators--I neglected to mention that. I don't know whether on every Banach space the closure of the diagonal operators contains the compact operators. $\endgroup$
    – Bill Johnson
    Commented Apr 9, 2015 at 13:34

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